Hey, guys. In this video, we're going to talk about these really incredible tools that we use when solving AC circuits called phasors. Alright. Let's get to it. Now a phasor is just a rotating vector. Okay. Phasor means phase vector. All the information contained by a phasor is contained in its x component. You can completely ignore the vertical components because it doesn't mean anything. Alright. Phasors are perfect for capturing all the information and representing it very easily for oscillating values like voltage and current, which we know oscillate. For instance, we know that the voltage is a function of time, looks like some maximum voltage times cosine of omega t. This is exactly what I've drawn here. I've given one cycle of voltage that undergoes a sinusoidal oscillation. Okay? And we want to look at how a phasor can easily represent this exact information. Now there are 4 times that I'm going to be interested in. What I'll call time 1, when the voltage is at a maximum and positive, time 2, when the voltage is okay? So these 4 diagrams here, okay? So these 4 diagrams here, sorry, t2, t3, t 4, are going to contain the phasor that represents the information about the voltage at each of those 4 times. These diagrams, by the way, are called phasor diagrams for obvious reasons. Okay? Now, initially, the voltage is at a maximum. In order for a phasor to be at its maximum, its entire length has to be along the x-axis. Okay. This is just because a vector that points along an axis, for instance, the x-axis, that's when that vector's component is largest. Okay? The x component of a vector is largest when that vector points along the x axis. Now the question is which side, left or right, do we want to put it on? By convention, to the right is considered positive and to the left is considered negative. So I'm going to draw the phasor like, negative. So I'm going to draw the phasor like this. It's entirely along the x axis which means that the voltage is at the largest value it could possibly be. Because the phasor, as it rotates, remember a phasor is a rotating vector, is not going to change length. Okay? So this is our voltage phasor. That's at time 1. Now at time 2, the voltage is 0. That means that it has to have no x component. So the phasor has to lie entirely along the vertical axis. The question is, is it up or is it down? These four diagrams that I have marked here, incidentally, are referred to as phasor diagrams for obvious reasons. Okay? Now, initially the voltage is at a maximum. To be at its maximum, a phasor must be fully extended along the x-axis. Okay. This merely indicates that when a vector is aligned with an axis, such as the x-axis, its respective component reaches its maximum value. Now the issue is which side, left or right, should we mount it? By convention, right is deemed positive and left negative. Thus, I will align the phasor like this—it lies entirely on the x-axis, signifying that the voltage has reached the highest possible value. Because the phasor rotates as a rotating vector, its length will not change. Okay? Hence, this represents our voltage phasor at time one. Now at time two, the voltage zeroes out. This necessitates a phasor devoid of any x-component, fully enclosed by the vertical axis. Does it point upwards or downwards? When devoid of any x-components at time two, we conventionally have the phasor pointing upwards. This standard assumes counterclockwise phasor rotation. Okay? At time three, the voltage returns to its maximum but inversely, making it negative. Thus, since it's at its fullest, the phasor should stretch entirely along the x-axis, but pointing leftwards because of its negativity. Okay. And finally, at t4, the voltage zeroes out again. Lacking any x-components, it must reside purely on the vertical axis. Initially directed leftward and rotating counterclockwise, it now faces downwards. And this constitutes the phasor, rotating counterclockwise with the same angular oscillation frequency omega. Simply put, if omega equals 2 per second, it completes two full rotations each second. Okay? Now phasors might seem odd initially as you encounter them. They require practice to be understood fully. So let's engage in an example to aid our familiarity with what a phasor entails. For this upcoming voltage phasor, is the voltage positive or negative? Recall, all relevant data lies along the x-axis. So that's all we focus on: the x-component. This. About as lengthy as this phasor. It's somewhat longer. Regardless. Since it points rightward, we identify this as positive. Okay? Its projection is termed such. It's projection onto the x-axis is positive. Okay? Now the reason phasors prove so invaluable, and why they are so heavily utilized, is that at any particular moment, if you pause time and capture a snapshot, phasors can be manipulated just as vectors are: they can be added, subtracted, and their magnitudes calculated through the Pythagorean theorem, exactly as you would determine a vector's magnitude. Let's demonstrate this with an example. In the subsequent phasor diagram, determine the net phasor's direction for the three displayed phasors. Is the resultant phasor quantity positive or negative? Assume, hypothetically, that these three phasors all describe voltage. Okay? Merely as an illustration—they could just as well delineate current, for example. What I must do is depict this as an entire net voltage phasor. Okay? Here we have two phasors facing identical directions. My apologies, I forgot to label these. V1, v2, and v3. V1 and v3 align along the same axis. Okay? Consequently, the resultant of these two will orient in v3's direction since v3 stretches longer. It resembles two forces aligned contrarily: the stronger force prevails. Thus, we maintain a phasor directed as v3 but slightly diminished in size. Now v2 stands alone because it's orthogonal. Here’s v2, here’s v3 minus v1. I'm uncertain which of these proves longer, v2 or v3 minus v1, but our net phasor will point somewhere in between here. Perhaps in this direction, maybe precisely along the axis, possibly below. In actuality, its precise alignment is irrelevant because it points right regardless. Our value remains invariably positive. Sorry, I shouldn't assert invariably. Our value will be positive. For the net phasor to yield a negative outcome, it would need to point leftward, which it clearly doesn’t. Okay? This only commences our exploration with phasors, gentlemen. Phasors can confuse just as vectors initially bewildered you, but with continual use, you’ll grow increasingly adept with phasors. In upcoming videos, we'll delve more into phasors within the specific contexts of voltage and current in circuits, crystallizing their functionality and application methods. Alright, guys. Thanks for watching.

# Phasors - Online Tutor, Practice Problems & Exam Prep

### Phasors

#### Video transcript

The following phasor diagram shows an arbitrary phasor during its first rotation. Assuming that it begins with an angle of 0° , if the phasor took 0.027 s to get to its current position, what is the angular frequency of the phasor?

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### Converting Between a Function and a Phasor

#### Video transcript

Hey guys, let's do a phasor example. In this case, an example that deals with relating the phasor's equation to the phasor diagram. Okay? The current in an AC circuit is given by this equation. Draw the phasor that corresponds to this current at 15 milliseconds, assuming the phasor begins at 0 degrees. Okay? So in the beginning, the phasor is going to start here at 0 degrees, and it's going to rotate through some amount of angle to arrive at its final position. We want to figure out what that angle is so we know where to draw the final position of this phasor. Remember that the angle is just ω, the angular frequency of the phasor times t. Okay? Now, we know that t is just 15 milliseconds. So what's the angular frequency of the phasor? Well, we're told that the angular frequency of the current is 377 rad/s, and that's going to be the angular frequency of the phasor. However, quickly it's oscillating on a function, right, if I were to draw this as an oscillating graph is going to be the same rate as how quickly it's oscillating on a phasor diagram. Those angular frequencies are the same. So this is just going to be 377 times 15 milliseconds, millis 10 to the negative 3, and this equals 5.66 radians. We want this to be in degrees because it's easiest to graph degrees for us or sorry to draw degrees on a diagram. Remember, you can convert by dividing this by π and multiplying it by 180 degrees. This is going to be 324 degrees. Okay, I'm going to minimize myself and draw this phasor diagram. It started from 0 degrees. Don't forget the phasor began at 0. So starting from 0 and rotating counterclockwise, this phasor ends up in the 4th quadrant because 324 is greater than 270 but less than 360. Okay? There are 2 other ways that you can represent this number if you want. This one is 324 as the full rotation. You can represent it from the negative y-axis if you want, and this would be 54 degrees, or you can represent it from the positive x-axis if you'd like, and this would be 36 degrees. Either way, this is correct, but the important thing to remember is that this value, 324, is how far it traveled from its initial position. It started at 0 degrees and phasors always rotate counterclockwise. So this is 324 degrees. Alright, guys. Thanks for watching.

An AC source oscillates with an angular frequency of 120 s^{-1} . If the initial voltage phasor is shown in the following phasor diagram, draw the voltage phasor after 0.01 s. (Select the correct absolute angle below of the phasor's location below after you have drawn it.)

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A phasor of length 4 begins at 0° . If it is rotating at ω = 250 s^{−1} , what is the value of the phasor after 0.007 s?